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Thread: Derivative of relative entropy

  1. #1
    Sep 2010

    (Solved) Derivative of relative entropy

    I am currently confused over relative entropy as it is defined here: Relative Entropy -- from Wolfram MathWorld
    Rewritten in a more tidy fashion, I have:
    \mathcal{H}(\Pr, {\Pr}_{0}) = \sum_{\omega \in \Omega} \Pr(\omega) \cdot \log\frac{\Pr(\omega)}{{\Pr}_{0}(\omega)}}

    I just computed the first and second derivative, because I want to demonstrate how a marginal alteration of probability \Pr(\hat\omega) will cause a growth of relative entropy that becomes the bigger, the more \Pr(\hat\omega) differs from {\Pr}_0(\hat\omega). At least this is what my textbook says on the intuition of relative entropy. Now I eneded up with this:

    \frac{\partial \mathcal{H}(\Pr,{\Pr}_{0})}{\partial \Pr(\hat\omega)} & = & \Big(\ln\frac{\Pr(\hat\omega)}{{\Pr}_{0}(\hat\omeg  a)} + 1\Big) \cdot \frac{1}{\ln 2}


    \frac{\partial^2 \mathcal{H}(\Pr,{\Pr}_{0})}{\partial \big(\Pr(\hat\omega)\big)^2} & = & \frac{1}{{\Pr}(\hat\omega)} \cdot \frac{1}{\ln 2}

    I am very much confused about the "+1" in the upper formula. Why is it there? Should setting \Pr(\hat\omega) = {\Pr}_0(\hat\omega) not cause a partial minimum? According to what I know, the derivative should then be zero, not \frac{1}{\ln 2}.

    Help is very appreciated!
    Best, Rafael
    Last edited by raphw; Mar 15th 2012 at 03:55 AM.
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  2. #2
    Sep 2010

    Re: Derivative of relative entropy

    I think, writing the problem down just answered my question. Of course, the partial derivative does not help me, since i the minimization must take the interdependency of all \omega \neq \hat\omega into account. By just deriving to \Pr(\omega) = 0, I would already find a minimum which is why my derivative indicates differently than my intention. Thanks for reading!
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